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victus00 [196]
3 years ago
14

A golf club experts an average force of 475 N for 0.006 seconds when it collides with a golf ball on a tee. What is the impulse

applied to the ball?
Physics
1 answer:
Ainat [17]3 years ago
8 0

Answer: 2.85 N.s

Explanation: Impulse is the product of force and time or expressed in the following formula:

J = F x t

= 475 N x 0.006 s

= 2.85 N.s

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Explanation:

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Solar collectors are parts of ______. a. passive solar heating systems b. active solar heating systems c. internal combustion en
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b. active solar heating systems

Explanation:

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You are looking at a yellow flower growing outside in the sunshine. Why does it look yellow?
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Answer:

It looks yellow because that is the only (major) color reflected.

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3 0
3 years ago
A vertical straight wire carrying an upward 28-A current exerts an attractive force per unit length of 7.83 X 10 N/m on a second
JulijaS [17]

Answer:

i_2 = 978750 A

Since the force between wires is attraction type of force so current must be flowing in upward direction

Explanation:

Force per unit length between two current carrying wires is given by the formula

F = \frac{\mu_0 i_1 i_2}{2 \pi d}

here we know that

F = 7.83 \times 10 N/m

d = 7.0 cm = 0.07 m

i_1 = 28 A

now we will have

F = \frac{4\pi \times 10^{-7} (28.0)(i_2)}{2\pi (0.07)}

7.83 \times 10 = \frac{2\times 10^{-7} (28 A)(i_2)}{0.07}

i_2 = 978750 A

Since the force between wires is attraction type of force so current must be flowing in upward direction

3 0
3 years ago
A toy of mass 0.190-kg is undergoing SHM on the end of a horizontal spring with force constant k = 350 N/m . When the toy is a d
vagabundo [1.1K]

Answer

a)0.0495 J

b)0.01681 m

c)0.7218 m/s

Explanation:

Given

Mass of the.toy M = 0.190 kg

force constant k = 350 N/m

Displacement from equilibrium x = 0.0140 m

Speed v = 0.400 m/s

a)What is the toy's total energy at any point of its motion?

The total energy at any point of it's motion can be calculated by adding together both the potential and kinetic energy of the toy, since it's posses potential energy when at rest and kinetic energy at motion

Total energy E = kinetic energy + potential energy

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.190)(0.4)² + ¹/₂ (350)(0.0140)²

E = 0.0495 J

Hence,the total energy is 0.0495 J

b) the amplitude of the motion can be calculated using below formula

Let amplitude = A

E = ¹/₂KA²

if we make Amplitude A the subject of the formula we have

A=√(2E/k)

But we have calculated our E up there, our K was given in question then if we substitute we have

A= √(2×0.0495)/350

Ans: 0.01681 m

Hence, our Amplitude is 0.01681 m

c) the the toy's maximum speed during its motion can be calculated using the expression below

Let maximum speed = vmax

E = (1/2)M * vmax^2

If we make vmax the subject of the formula we have

vmax =√(2E/m)

vmax= √(2×0.0495)/0.190

vmax=0.7218 m/s

Hence our vmax is 0.7218 m/s

8 0
3 years ago
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